使用std::string声明一个std::bitset数组

Question

我目前正在尝试声明一个由17个32位std::bitsets数组组成的数组。我这样做：

std::bitset<32> mTestInstruction[17]
{
    std::string("01000000001000000000000000000001"),
    std::string("01000000011000000000000001100011"),
    std::string("01000000101000000000000000000001"),
    std::string("10100000000000000000000000001010"),
    std::string("00000000100000010000000010000010"),
    std::string("00000000110001010010000000000001"),
    std::string("01001000111001010000000000000000"),
    std::string("01000100001000110000000000000011"),
    std::string("01000000001000010000000000000001"),
    std::string("10000000000000000000000000000011"),
    std::string("00000000010000000000000000000001"),
    std::string("00000000111000000000000000000001"),
    std::string("00000000111001110000100000000001"),
    std::string("01000000010000100000000000000001"),
    std::string("01000100001000100000000000000010"),
    std::string("10000000000000000000000000001100"),
    std::string("11100000000000000000000000001000"),
};

我收到了以下错误：

错误：无法将 'std :: __ cxx11 :: basic_string < char> (((const char *)" 01000000001000000000000000000001"), std :: allocator <char>（）)'从 'std :: __ cxx11 :: string {aka std :: __ cxx11 :: basic_string <char>}'转换为 'std :: bitset <32u>'

对于每个比特串，我不理解为什么会出现这种情况，因为根据cpp参考，std :: string是构建bitset的有效方法。

请问有谁能指出如何解决这个问题吗？

- SaucyPandora

---

你说得对：确实有一个接受 std::string 的构造函数。不幸的是，它是 explicit 的。 - max66

Answer 1

您需要像这样调用 std::bitset 的构造函数：

std::bitset< 32 > mTestInstruction[17]
{
    std::bitset< 32 >( std::string( "01000000001000000000000000000001" ) ),
    std::bitset< 32 >( std::string( "01000000011000000000000001100011" ) ),
    // ...
};

甚至更短：

std::bitset< 32 > mTestInstruction[17]
{
    std::bitset< 32 >( "01000000001000000000000000000001" ),
    std::bitset< 32 >( "01000000011000000000000001100011" ),
    // ...
};

您的代码无法运行的原因是因为std::bitset的接受std::string的构造函数被标记为显式(explicit) (参见这里)。

从 C++14 开始，您可以使用二进制字面值(binary literals)，如下所示:

std::bitset< 32 > mTestInstruction[17]
{
    0b01000000001000000000000000000001ull,
    0b01000000011000000000000001100011ull,
    // ...
};

- NutCracker

---

@GManNickG的回答已经编辑完毕。感谢您的帮助。 - NutCracker

Answer 2

你可以这样做：

std::bitset<32> mTestInstruction[17]{
        std::bitset<32>{std::string("01000000001000000000000000000001")},
        std::bitset<32>{std::string("01000000011000000000000001100011")},
        std::bitset<32>{std::string("01000000101000000000000000000001")},
        std::bitset<32>{std::string("10100000000000000000000000001010")},
        std::bitset<32>{std::string("00000000100000010000000010000010")},
        std::bitset<32>{std::string("00000000110001010010000000000001")},
        std::bitset<32>{std::string("01001000111001010000000000000000")},
        std::bitset<32>{std::string("01000100001000110000000000000011")},
        std::bitset<32>{std::string("01000000001000010000000000000001")},
        std::bitset<32>{std::string("10000000000000000000000000000011")},
        std::bitset<32>{std::string("00000000010000000000000000000001")},
        std::bitset<32>{std::string("00000000111000000000000000000001")},
        std::bitset<32>{std::string("00000000111001110000100000000001")},
        std::bitset<32>{std::string("01000000010000100000000000000001")},
        std::bitset<32>{std::string("01000100001000100000000000000010")},
        std::bitset<32>{std::string("10000000000000000000000000001100")},
        std::bitset<32>{std::string("11100000000000000000000000001000")},
    };

- Ari

---

这似乎只是重复了已接受的答案，没有添加任何内容。 - scohe001